Theorem disjeq1f

Description: Equality theorem for disjoint collection. (Contributed by Mario Carneiro, 14-Nov-2016)

	Ref	Expression
Hypotheses	disjss1f.1	$$⊢\underset{\_}{Ⅎ}xA$$
	disjss1f.2	$$⊢\underset{\_}{Ⅎ}xB$$
Assertion	disjeq1f	$$⊢A=B\to \left(\underset{x\in A}{Disj}C\leftrightarrow \underset{x\in B}{Disj}C\right)$$

Proof

Step	Hyp	Ref	Expression
1		disjss1f.1	$$⊢\underset{\_}{Ⅎ}xA$$
2		disjss1f.2	$$⊢\underset{\_}{Ⅎ}xB$$
3		eqimss2	$$⊢A=B\to B\subseteq A$$
4	2 1	disjss1f	$$⊢B\subseteq A\to \left(\underset{x\in A}{Disj}C\to \underset{x\in B}{Disj}C\right)$$
5	3 4	syl	$$⊢A=B\to \left(\underset{x\in A}{Disj}C\to \underset{x\in B}{Disj}C\right)$$
6		eqimss	$$⊢A=B\to A\subseteq B$$
7	1 2	disjss1f	$$⊢A\subseteq B\to \left(\underset{x\in B}{Disj}C\to \underset{x\in A}{Disj}C\right)$$
8	6 7	syl	$$⊢A=B\to \left(\underset{x\in B}{Disj}C\to \underset{x\in A}{Disj}C\right)$$
9	5 8	impbid	$$⊢A=B\to \left(\underset{x\in A}{Disj}C\leftrightarrow \underset{x\in B}{Disj}C\right)$$